STRUCTURE OF Cl-35 AND Ar-35
.By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of the stable Cl-35 with S = +3/2 ''' In the diagram of Cl-35 you can see the core of Cl-35 which consists of the parallelepiped of Mg-24 (Six horizontal squares and planes from p1 to n12 with S=0) and of two alpha particles. Especially at the central parallelepiped of the structure of Be-8 we add the first alpha particle of p13, n13, p14 and n14 and the second alpha particle of p15, n15, p16 and n16. Since the alpha particles have S=0 one concludes that the core (from p1 to n16) has S=0 .The advantage of this core with two alpha particles is that it forms blank positions in order to receive one proton and two neutrons for making the stable Cl-35 with 17 protons and 18 neutrons with a spin S = +3/2 which gives the total spin S = +3/2 . In such additional nucleons at the blank positions of p17 and n17 there exist the deuteron with p17 and n17. It makes stable pn and np bonds because at points p17 and n17 one observes three pn and two np bonds respectively. For example at p17 of the fifth horizontal plane (+HP5) we observe the bonds (p17-n10) the (p17-n17) and the p17-n14). Also at point n17 we observe the tree np bonds as the ( n17-p9), the (n17-p17) and the ( n17-p17). Of course this bond of ( n17-p17) has not the structure of an isolated deuteron because it is connected with the neighbor protons . It has a spin S= +1 because it belongs to the fifth horizontal square with nucleons of positive spins. Note that here the nucleons of the first horizontal square of Mg-24 have positive spins while the nucleons of the sixth horizontal square have negative spins. On the other hand the additional n18 makes the two np bonds like (n18-p10) and (n18-p16) which lead to the stability of Cl-35. Similarly the additional n18 has spin S = +1/2 because it belongs to the same fifth horizontal square with nucleons of positive spins. Thus the total spin of Cl-35 is S = +1 +1/2 = +3/2. '''Structure of the unstable Ar-35 with S = +3/2 In the diagram of Cl-35 one sees that the Cl-35 has an additional stable deuteron and an additional n18 which makes the two np bonds per nucleon able to overcome the nn repulsions of short range. However in the structure of Ar-35 one observe the same additional structure of one deuteron but with an additional proton . Thus the additional n18 must be replaced by an additional proton. Since Ar-35 consists of 18 protons and 17 neutrons with the same S = +3/2 one concludes that the additional proton will be at a blank position between the neutrons n9 and n16 with the same positive spin S = +1/2 of that of n18, because it belongs to the same fifth horizontal square of positive spins. On purpose the position of p18 is not shown here because it is similar to the position of n18 in which the neutron was between the two protons p10 and p16. Thus in the absence of n18 the additional p18 makes only the two pn bonds like (p18-n9) and (p18-n16) which are unable to overcome the strong repulsive energy Q of the pp repulsions of long range. Under this condition the p18 turns to the neutron n18 with the two np bonds which can overcome the nn repulsions of short range. That is p18 + Q = n18 + positron + neutrino In other words the unstable Ar-35 with 18 protons and 17 neutrons turns into the stable Cl-35 with 17 protons and 18 neutrons. ' ' ' ' ' Cl-35 with S = +3/2 ' ' ' ' ' '' p12............n12' ' ' '' p11............n11' ''' -HP6''' ' p17...............n10............p10...........n18 ' ' n17............... p9...............n9 +HP5 ' ' n14...............p8...............n8...............p16 ' ' p14...............n7................p7...............n16 -HP4 ' ' p13...............n6................p6...............n15 ' ' n13............p5...............n5...............p15 +HP3 ' ' p4...............n4 ' ' n3...............p3 -HP2 ' ' n2...............p2 ' ' p1...............n1 +HP1' ' ''' '' '' ' ''' Category:Fundamental physics concepts